题解 | 公共子串计算

#include <iostream>
#include <string>

using namespace std;

#define N 200
int dp[2][N];

int getLongestCommonSubString(string& a, string& b) {
    int lena = a.size() - 1;
    int lenb = b.size() - 1;
    int maxlen = 0;
    for (int i = 1; i <= lena; ++i) {
        for (int j = 1; j <= lenb; ++j) {
            if (a[i] == b[j]) {
                dp[i % 2][j] = dp[(i - 1) % 2][j - 1] + 1;
                if (maxlen < dp[i % 2][j]) maxlen = dp[i % 2][j];
            } else {
                dp[i % 2][j] = 0; // 清零,防止错误数据积累
            }
        }
    }
    return maxlen;
}

int main() {
    string s1;
    string s2;
    cin >> s1 >> s2;
    s1 = " " + s1;
    s2 = " " + s2;
    cout << getLongestCommonSubString(s1, s2) << endl;

    return 0;
}